Ans: C
Let $a$ and $r$ be the first term and the common ratio of the geometric sequence. Then we have
Let $a$ and $r$ be the first term and the common ratio of the geometric sequence. Then we have
$\left\{ \begin{array}{ll}
a + ar = 8 & \ldots \unicode{x2460} \\
ar^2 = 18 & \ldots \unicode{x2461}
\end{array} \right.$
$\unicode{x2461} \div \unicode{x2460}$, we have
$\begin{array}{rcl}
\dfrac{ar^2}{a(1+r)} & = & \dfrac{18}{8} \\
4r^2 & = & 9(1+r) \\
4r^2 – 9r – 9 & = & 0 \\
(r-3)(4r+3) & = & 0 \\
\end{array}$
Therefore, $r=3$ or $r = \dfrac{-3}{4}$. Sub. $r=3$ into $\unicode{x2461}$, we have
$\begin{array}{rcl}
a(3)^2 & = & 18 \\
a & = & 2
\end{array}$
Sub. $r=\dfrac{-3}{4}$ into $\unicode{x2461}$, we have
$\begin{array}{rcl}
a(\dfrac{-3}{4})^2 & = & 18 \\
a & = & 32
\end{array}$
Therefore, the first term of the sequence is $2$ or $32$.
